In an isosceles triangle ABC, the bisectors AL and BM of equal angles A and B intersect at point O.

In an isosceles triangle ABC, the bisectors AL and BM of equal angles A and B intersect at point O. Find the angle of the BMC (in degrees) if the outer angle C of the triangle is 100 °.

The angle ACB, by condition, is equal to 100, then the angle ACB adjacent to it is equal to 180 – 100 = 80.

Since the triangle ABC is isosceles, the angle ABC = BAC = (180 – ACB) / 2 = (180 – 80) / 2 = 50.

Since, by condition, BM is the bisector of the angle, then the angle СBМ = MBА = ABC / 2 = 50/2 = 25.

Then in the BCM triangle, the angle BMC = 180 – ACB – СBM = 180 – 80 – 25 = 75.

Answer: The BCM angle is 75.